Formula to calculate facing time in a lathe?

[mattslay]

Does anyone know the formula to calculate the theoretical time to make a facing cut across a part (in a CNC lathe)?

For instance...

Imagine a ring that is 20" O.D. and has a 4" I.D., and I want to estimate the time to make 1 facing cut across the face.

(I'm looking for the formula, not just the answer to this sample set of numbers.)

O.D. = 20
I.D. = 4
Feedrate = .010" per rev (constant)
SFM = 300 (spindle speed will vary to hold this number)



Matt Slay
Jordan Machine Co.
Birmingham, AL

[Geof]

I don't know of a single formula but you should be able to calculate it this way:

You know what the rpm is to start; 20" dia., 300sfm = ? rpm this is an easy calculation.

You have the spindle speed clamped so at some diameter you reach the clamp rpm so you do the calculation backward to find what diameter the tool is at.

Now take the average of the starting rpm and the clamp rpm and take the average of the starting diameter and the diameter at which the clamp rpm is reached.

Use these two averages to calculate the time taken to go from the start to to the diameter at which the speed gets clamped.

Once the spindle is running at a constant speed it is easy to calculate the remaining time to reach the 4" diameter.

[Eurisko]

Matt,

This is classic Calculus. It's been over 25 years since I took it, but I was able to refresh my memory with a "Calculus for Dummies" book.

Feedrate is a function of RPM, which is a function of the radius (and CSS) when facing.

Time (min/inch) is the inverse of feedrate (in/min)

So to make a long story short, here is the general equation:

F=feedrate in inches per revolution
R1=starting radius in inches (larger radius)
R2=ending radius in inches (smaller radius)
CSS=constant surface speed in feet per minute
PI=PI

total time in minutes = (PI / ( F X CSS X 12)) X (R1 ^ 2 - R2 ^2)

Plug your numbers into this equation and you'll get 8.3776 minutes.
The answer is exact. I rounded it to 4 decimal places.

If you clamp your spindle speed before reaching maximum rpm's, (at 4 inches) it will take longer.

[PixMan]

Eurisko,

Your formula is perfect except for the missing element of maximum spindle speed clamp value.

[mattslay]

Well, I got several suggestions to figure the RPM at the average dia and then use the feedrate over the distance to be traveled. I did not like this intuitive approach, since my quest was for an exact answer (within a second or two at least), and I was sure averaging was not going to get that accurate . So, I wrote a C# program to iterate over the face at the feedrate amount, and calculate the rpm at each increment, and then determine the time for that rev to occur, and then sum them all up (kind of sounds like Calculus, huh, Eurisko?). (See my C# code below)

As it turns out, the avereage and the calculus are EXACTLY the same answer as each other, and my iterative solution also gives (essentially) the same answer (within 1/2 second).

Results:

Avereage approach: 8.3776 min
Calculus formula: 8.3776
Iterative loop: 8.3705 min

Anyway, I am conviced that the intuitive "average" model actually gives the correct answer, as well as the beautiful calculus formula, as well as my C# hacking.

Agree?

(Oh yeah, and, none of these account for the clamping out of the spindle speed as someone mentioned, but that would be easy to add to my iterative loop, it's just a test to limit rpm after calculating it)

Here is my C# code, in case anyone wants review it, or play around with it in Visual Studio:



using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;

namespace ConsoleApplication1
{
class Program
{

static void Main(string[] args)
{
FacingCuts Cut = new FacingCuts();
Cut.InitialDia = 20;
Cut.FinishDia = 4;
Cut.SFM = 300;
Cut.FeedRate = 0.010;

Console.WriteLine("Cycle Time {0} min.", Cut.CalculateFacingTime());
Console.ReadLine();

}
}


class FacingCuts
{
public double InitialDia = 0;
public double FinishDia;
public double SFM;
public double FeedRate;

public double CalculateFacingTime()
{
int x;
double CurrentDia;
double RPM;
double Circumference;
double TotalTime = 0;
double FaceDistance = Math.Abs(InitialDia - FinishDia)/2.0;
if (InitialDia > FinishDia)
{
FeedRate = FeedRate * -1;
}
int NumberOfRevs = Math.Abs((int)(FaceDistance / FeedRate));

for (x = 1; x <= NumberOfRevs; x++)
{
CurrentDia = InitialDia + (x * 2 * FeedRate);
Circumference = CurrentDia * Math.PI / 12;
RPM = SFM / Circumference;
TotalTime = TotalTime + (1 / RPM);
}

return TotalTime;
}


}

}

[Eurisko]

Matt,

I'm not well versed in C#, I tested the Calculus equation with QBasic.

An iterative approach is the way to go if you have to deal with clamping the spindle speed.

In reality, there are four possibilities.
1) The spindle speed is clamped lower than the starting RPM
2) The spindle speed is clamped at the starting RPM
3) The spindle speed is clamped between the starting and ending RPM's
4) The spindle speed is clamped at or higher than the ending RPM

An interesting approach is to calculate how much time it takes to reach the clamp RPM.

CLAMP=clamp RPM

time in minutes = (PI / ( F X CSS X 12)) X ( R1 ^2 - ( (CSS X 6) / ( PI X CLAMP)) ^2 )

If the clamp time is negative, condition 1 is met
If the clamp time is zero, condition 2 is met
If the clamp time is greater than zero and less than the actual time, condition 3 is met.
If the clamp time is greater than or equal to the actual time, condition 4 is met.

Of course, you could always calculate the clamp radius, which appears in the equation as (CSS X 6) /(PI X CLAMP)

There's just something about "negative time" that appeals to me.

[Changedsoul]

This is great, I have just recently been trying to figure this out my self. Thanks for the nice simple formula that works out to be very accurate.

I was also wondering how you could calculate the time needed to move along a radius. Would it be the same principle? Because now your moving in two axis and not one.

I was thinking about finding the average diameter and take the spindle speed at that point, then calculate the arc length the tool would move on, and use that length with that spindle speed with a given feed rate to produce the time.

Would this be correct? And what about the spindle clamp speed as well?

[BlueChip]

Why bother ... KipwareCYC ... www.KentechInc.com ... will do it all for you.

[Changedsoul]

Well yes there is always that, but there is a slight problem for me, This software cost allot of money. Most of all though, I am actually trying to write a program to calculate the time it takes to machine a part. It scans through the prog adding up all the movments, but also adds time taken for moving steady rest and tail stock and all other good time consuming things up for a grand total.

I work at a place where we have a programming department that gives us the programs and how long it should take to run the part, but more than not, they have the times completely wrong.

So as a way for me to learn C++ like I wanted to I figured I would have a go at this. I just dont know what i can do to calculate radius movements. I run large parts so a radius is a big factor. It could take 1 min to move along a small radius.

So anyways...thats why I cannot just use that software, also I dont have hundreds of dollars in my wallet

[donl517]

Hey Matt,

Thanks for the great example, Have you done any other work on this?

Don